2025 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Find the highest common factor of the numerical coefficients \(18\) and \(12\), which is \(6\).
Find the highest common factor of the variable parts \(x^2y\) and \(xy^2\), which is \(xy\).

Factor out \(6xy\):
\(18x^2y - 12xy^2 = 6xy(3x - 2y)\)

B2 for \(6xy(3x - 2y)\)
OR
B1 for any correct partial factorisation with at least two common factors taken out, e.g. \(6(3x^2y - 2xy^2)\), \(3xy(6x - 4y)\), \(2xy(9x - 6y)\), \(xy(18x - 12y)\), or \(6xy(\text{two-term algebraic expression})\)

Find the size of one exterior angle:
\(\text{Exterior angle} = 180^\circ - 156^\circ = 24^\circ\)

The sum of the exterior angles of any convex polygon is \(360^\circ\):
\(\text{Number of sides} = \frac{360^\circ}{24^\circ} = 15\)

M1 for \(180 - 156\) or \(\frac{(n - 2) \times 180}{n} = 156\) oe
A1 for 15

Multiply the numbers and powers of 10 separately:
\((4.5 \times 8) \times (10^7 \times 10^{-3}) = 36 \times 10^{7 + (-3)} = 36 \times 10^4\)

Convert \(36 \times 10^4\) into standard form \(A \times 10^n\) where \(1 \le A < 10\):
\(36 \times 10^4 = (3.6 \times 10^1) \times 10^4 = 3.6 \times 10^5\)

B2 for \(3.6 \times 10^5\)
OR
M1 for \(36 \times 10^4\) or \(360\,000\) or \(3.6 \times 10^k\) (where \(k \neq 5\))

Rearrange the given equation into the form \(y = mx + c\):
\(y = 3x - 2.5\)

The gradient of the given line is \(m_1 = 3\).

For perpendicular lines, \(m_1 \times m_2 = -1\):
\(m_2 = -\frac{1}{3}\)

M1 for gradient of given line is 3, soi, or for using \(m_1 m_2 = -1\) with their identified gradient
A1 for \(-\frac{1}{3}\) oe

Divide the numerical coefficients:
\(4.8 \div 1.5 = 3.2\)

Divide the powers of 10:
\(10^7 \div 10^{-3} = 10^{7 - (-3)} = 10^{10}\)

Combine into standard form:
\(3.2 \times 10^{10}\)

B2 for \(3.2 \times 10^{10}\)
OR
B1 for 3.2 or \(10^{10}\) seen in intermediate working, or for \(32 \times 10^9\) or \(0.32 \times 10^{11}\)

Divide the numerical coefficients and subtract the powers of 10:
\((3.6 \div 8) \times 10^{7 - 2} = 0.45 \times 10^5\)

Convert to standard form:
\(0.45 \times 10^5 = 4.5 \times 10^4\)

M1 for \(0.45 \times 10^5\) or \(45\,000\) seen
A1 for \(4.5 \times 10^4\) cao

Find the highest common factor of the terms \(15px\) and \(10qx\):
\(\text{HCF}(15, 10) = 5\)
Common variable factor is \(x\).
So the common factor is \(5x\).

Factor out \(5x\):
\(15px - 10qx = 5x(3p - 2q)\)

B2 for \(5x(3p - 2q)\)
or B1 for partial factorisation: \(5(3px - 2qx)\) or \(x(15p - 10q)\)

Find the lowest common denominator of 12 and 9, which is 36.
Convert both fractions to have denominator 36:
\(\dfrac{7 \times 3}{12 \times 3} = \dfrac{21}{36}\)
\(\dfrac{2 \times 4}{9 \times 4} = \dfrac{8}{36}\)

Subtract the numerators:
\(\dfrac{21}{36} - \dfrac{8}{36} = \dfrac{13}{36}\)

M1 for writing both fractions over a correct common denominator, e.g. \(\dfrac{21}{36} - \dfrac{8}{36}\) or \(\dfrac{63}{108} - \dfrac{24}{108}\)
A1 for \(\dfrac{13}{36}\) cao

Each exterior angle is \(180^\circ - 156^\circ = 24^\circ\).

The sum of exterior angles of any convex polygon is \(360^\circ\).
Number of sides \(n = \dfrac{360^\circ}{24^\circ} = 15\).

M1 for \(180 - 156\) or \((n - 2) \times 180 = 156n\) oe
A1 for 15 cao

Apply the exponent \(\frac{2}{3}\) to each component inside the bracket:
\(64^{\frac{2}{3}} = (\sqrt[3]{64})^2 = 4^2 = 16\)
\((w^{12})^{\frac{2}{3}} = w^{12 \times \frac{2}{3}} = w^8\)

Combining these gives \(16w^8\).

B2 for \(16w^8\)
or B1 for \(16w^k\) (where \(k \neq 8\)) or \(cw^8\) (where \(c \neq 16\)) or \(64^{2/3} = 16\)

(a) A quadrilateral with four equal sides and perpendicular diagonals of equal length is a square.

(b) A regular polygon with \(n\) sides has \(n\) lines of symmetry. For a regular hexagon, \(n = 6\).

(c) A general parallelogram has rotational symmetry of order 2 (it maps onto itself after a rotation of \(180^\circ\) and \(360^\circ\)).

(a) B1 for square
(b) B1 for 6
(c) B1 for 2

Convert \(24\text{ minutes}\) into hours:
\[ 24\text{ minutes} = \frac{24}{60}\text{ h} = 0.4\text{ h} = \frac{2}{5}\text{ h} \]

Set up the time difference equation:
\[ \frac{36}{x - 3} - \frac{36}{x} = \frac{2}{5} \]

Multiply through by \(5x(x - 3)\):
\[ 5 \times 36x - 5 \times 36(x - 3) = 2x(x - 3) \]
\[ 180x - 180x + 540 = 2x^2 - 6x \]
\[ 2x^2 - 6x - 540 = 0 \]
\[ x^2 - 3x - 270 = 0 \]

Factorise or use the quadratic formula:
\[ (x - 18)(x + 15) = 0 \]

Since speed must be positive, \(x = 18\).

The average speed on the outward journey is \(18\text{ km/h}\).

M1 for \(\frac{36}{x - 3} - \frac{36}{x} = \frac{24}{60}\) oe
M1 for correctly clearing denominators, e.g. \(36x - 36(x - 3) = 0.4x(x - 3)\) or \(x^2 - 3x - 270 = 0\)
M1 for factorising \((x - 18)(x + 15) = 0\) or correct substitution into the quadratic formula for their 3-term quadratic
A1 for \(18\) cao (must reject \(-15\))

The overall dimensions including the \(1\text{ m}\) path on all sides are:
Length \(= (2x + 5) + 2(1) = 2x + 7\)
Width \(= (x + 2) + 2(1) = x + 4\)

Set up the area equation:
\[ (2x + 7)(x + 4) = 98 \]
\[ 2x^2 + 8x + 7x + 28 = 98 \]
\[ 2x^2 + 15x - 70 = 0 \]

Apply the quadratic formula:
\[ x = \frac{-15 \pm \sqrt{15^2 - 4(2)(-70)}}{2(2)} \]
\[ x = \frac{-15 \pm \sqrt{225 + 560}}{4} \]
\[ x = \frac{-15 \pm \sqrt{785}}{4} \]

Since \(x > 0\):
\[ x = \frac{-15 + 28.01785...}{4} \approx 3.25446... \]

Rounding to 3 significant figures gives \(x = 3.25\).

M1 for setting up the total dimensions as \((2x + 7)\) and \((x + 4)\) soi
M1 for expanding and forming 3-term quadratic \(2x^2 + 15x - 70 = 0\) oe
M1 for correct use of quadratic formula: \(x = \frac{-15 \pm \sqrt{15^2 - 4(2)(-70)}}{2(2)}\) or \(\frac{-15 \pm \sqrt{785}}{4}\)
A1 for \(3.25\) or \(3.254...\) (reject negative solution)

Let \(n\) be the original number of students.
Original cost per person \(= \frac{240}{n}\)
New cost per person \(= \frac{240}{n + 4}\)

Set up the equation representing the difference in cost:
\[ \frac{240}{n} - \frac{240}{n + 4} = 3 \]

Multiply through by \(n(n + 4)\):
\[ 240(n + 4) - 240n = 3n(n + 4) \]
\[ 240n + 960 - 240n = 3n^2 + 12n \]
\[ 3n^2 + 12n - 960 = 0 \]

Divide the entire equation by \(3\):
\[ n^2 + 4n - 320 = 0 \]

Factorise the quadratic:
\[ (n + 20)(n - 16) = 0 \]

Since the number of students must be positive, \(n = 16\).

The original number of students is \(16\).

M1 for \(\frac{240}{n} - \frac{240}{n + 4} = 3\) oe
M1 for multiplying by common denominator to obtain \(240(n + 4) - 240n = 3n(n + 4)\) oe
M1 for simplifying to \(n^2 + 4n - 320 = 0\) and attempting to solve (e.g. \((n + 20)(n - 16) = 0\))
A1 for \(16\) cao

Using Pythagoras' theorem:
\[ (x - 1)^2 + (2x + 2)^2 = (2x + 3)^2 \]

Expand the terms:
\[ (x^2 - 2x + 1) + (4x^2 + 8x + 4) = 4x^2 + 12x + 9 \]
\[ 5x^2 + 6x + 5 = 4x^2 + 12x + 9 \]

Rearrange into standard quadratic form:
\[ x^2 - 6x - 4 = 0 \]

Solve using the quadratic formula:
\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-4)}}{2(1)} \]
\[ x = \frac{6 \pm \sqrt{36 + 16}}{2} = \frac{6 \pm \sqrt{52}}{2} = 3 \pm \sqrt{13} \]

Since side length \(x - 1 > 0\), \(x > 1\):
\[ x = 3 + \sqrt{13} \approx 3 + 3.60555... = 6.60555... \]

Rounding to 3 significant figures gives \(x = 6.61\).

M1 for applying Pythagoras' theorem: \((x - 1)^2 + (2x + 2)^2 = (2x + 3)^2\)
M1 for correctly expanding terms: \(x^2 - 2x + 1 + 4x^2 + 8x + 4 = 4x^2 + 12x + 9\)
M1 for reducing to \(x^2 - 6x - 4 = 0\) and applying the quadratic formula \(\frac{6 \pm \sqrt{(-6)^2 - 4(1)(-4)}}{2}\)
A1 for \(6.61\) or \(6.605...\) or \(3 + \sqrt{13}\) (reject negative solution)

The base of the tray has:
Length \(= 30 - 2x\)
Width \(= 20 - 2x\)

Set up the area equation:
\[ (30 - 2x)(20 - 2x) = 264 \]
\[ 600 - 60x - 40x + 4x^2 = 264 \]
\[ 4x^2 - 100x + 336 = 0 \]

Divide by \(4\):
\[ x^2 - 25x + 84 = 0 \]

Factorise:
\[ (x - 21)(x - 4) = 0 \]

Since the width of the sheet is \(20\text{ cm}\), \(2x < 20\), so \(x < 10\).
Therefore, \(x = 4\).

M1 for base dimensions \((30 - 2x)\) and \((20 - 2x)\) seen or used
M1 for \((30 - 2x)(20 - 2x) = 264\)
M1 for simplifying to \(4x^2 - 100x + 336 = 0\) or \(x^2 - 25x + 84 = 0\) and attempting to solve
A1 for \(4\) cao (must discard \(x = 21\) with valid justification or final single answer)

(a) First find the diagonal of the base \(AC\):
\(AC^2 = AB^2 + BC^2 = 14^2 + 10^2 = 196 + 100 = 296\)
\(AC = \sqrt{296} \approx 17.205\text{ cm}\)

Since \(M\) is the centre of the base:
\(AM = \frac{1}{2}AC = \frac{\sqrt{296}}{2} = \sqrt{74} \approx 8.6023\text{ cm}\)

In right-angled triangle \(VMA\):
\(VM^2 = VA^2 - AM^2 = 13^2 - 74 = 169 - 74 = 95\)
\(VM = \sqrt{95} \approx 9.7468 \approx 9.75\text{ cm}\)

(b) In right-angled triangle \(VMA\), the angle of elevation is \(\text{angle } VAM\):
\(\cos(\text{angle } VAM) = \frac{AM}{VA} = \frac{\sqrt{74}}{13} \approx 0.6617\)
\(\text{angle } VAM = \cos^{-1}(0.6617) \approx 48.568^\circ \approx 48.6^\circ\)

(a)
M1 for \(14^2 + 10^2\) or \(7^2 + 5^2\) soi
M1 for \(13^2 - (\text{their } AM)^2\)
A1 for 9.75 or 9.746 to 9.747 or \(\sqrt{95}\)

(b)
M1 for \(\cos(VAM) = \frac{\text{their } AM}{13}\) or \(\sin(VAM) = \frac{\text{their } VM}{13}\) or \(\tan(VAM) = \frac{\text{their } VM}{\text{their } AM}\)
A1 for 48.6 or 48.56 to 48.57

(a) Find angle \(AHB\):
\(\text{angle } AHB = 138^\circ - 64^\circ = 74^\circ\)

Apply the cosine rule in triangle \(HAB\):
\(AB^2 = HA^2 + HB^2 - 2(HA)(HB)\cos(74^\circ)\)
\(AB^2 = 45^2 + 72^2 - 2(45)(72)\cos(74^\circ)\)
\(AB^2 = 2025 + 5184 - 6480\cos(74^\circ) = 7209 - 1786.131 = 5422.869\)
\(AB = \sqrt{5422.869} \approx 73.640 \approx 73.6\text{ km}\)

(b) Area of triangle \(HAB\):
\(\text{Area} = \frac{1}{2} \times 45 \times 72 \times \sin(74^\circ) = 1620 \times \sin(74^\circ) \approx 1557.244\text{ km}^2\)

Let \(d\) be the shortest distance from \(H\) to \(AB\):
\(\frac{1}{2} \times AB \times d = \text{Area}\)
\(d = \frac{2 \times 1557.244}{73.640} \approx 42.293 \approx 42.3\text{ km}\)

(a)
B1 for \(\text{angle } AHB = 74^\circ\) soi
M1 for \(45^2 + 72^2 - 2(45)(72)\cos(74)\)
A1 for 73.6 or 73.64...

(b)
M1 for \(0.5 \times 45 \times 72 \times \sin(74) = 0.5 \times (\text{their } AB) \times d\) oe
A1 for 42.3 or 42.29 to 42.30

(a) In the horizontal rectangular base \(ABCD\):
\(AC^2 = AD^2 + DC^2 = 9^2 + 16^2 = 81 + 256 = 337\)
\(AC = \sqrt{337} \approx 18.358\text{ cm}\)

Since \(AE\) is vertical and perpendicular to the base \(ABCD\), triangle \(EAC\) is right-angled at \(A\):
\(EC^2 = AC^2 + AE^2 = 337 + 7^2 = 337 + 49 = 386\)
\(EC = \sqrt{386} \approx 19.647 \approx 19.6\text{ cm}\)

(b) The angle between \(EC\) and the base \(ABCD\) is \(\text{angle } ECA\):
\(\tan(\text{angle } ECA) = \frac{AE}{AC} = \frac{7}{\sqrt{337}} \approx 0.3813\)
\(\text{angle } ECA = \tan^{-1}(0.3813) \approx 20.875^\circ \approx 20.9^\circ\)

(a)
M1 for \(9^2 + 16^2\) or \(AC = \sqrt{337}\) or \(18.35...\)
M1 for \((\text{their } AC)^2 + 7^2\) or \(9^2 + 16^2 + 7^2\)
A1 for 19.6 or 19.64 to 19.65 or \(\sqrt{386}\)

(b)
M1 for \(\tan(ECA) = \frac{7}{\text{their } AC}\) or \(\sin(ECA) = \frac{7}{\text{their } EC}\) or \(\cos(ECA) = \frac{\text{their } AC}{\text{their } EC}\)
A1 for 20.9 or 20.87 to 20.88

(a) Use the cosine rule in triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(112^\circ)\)
\(AC^2 = 85^2 + 62^2 - 2(85)(62)\cos(112^\circ)\)
\(AC^2 = 7225 + 3844 - 10540\cos(112^\circ)\)
\(AC^2 = 11069 - 10540(-0.3746) = 11069 + 3948.35 = 15017.35\)
\(AC = \sqrt{15017.35} \approx 122.545 \approx 123\text{ m}\)

(b) Total area \(= \text{Area}(\triangle ABC) + \text{Area}(\triangle ACD)\)
\(\text{Area}(\triangle ABC) = \frac{1}{2} \times 85 \times 62 \times \sin(112^\circ) = 2635 \times \sin(112^\circ) \approx 2443.13\text{ m}^2\)
\(\text{Area}(\triangle ACD) = \frac{1}{2} \times 122.545 \times 78 \times \sin(42^\circ) \approx 3197.95\text{ m}^2\)
\(\text{Total Area} = 2443.13 + 3197.95 = 5641.08 \approx 5640\text{ m}^2\)

(a)
M1 for \(85^2 + 62^2 - 2(85)(62)\cos(112)\)
A1 for 15017 to 15020
A1 for 123 or 122.5 to 122.6

(b)
M1 for \(0.5 \times 85 \times 62 \times \sin(112) + 0.5 \times (\text{their } AC) \times 78 \times \sin(42)\) oe
A1 for 5640 or 5641 to 5642

(a) First, find position vectors \(\vec{OP}\) and \(\vec{OQ}\):
Since \(OP : PA = 3 : 1\), \(\vec{OP} = \frac{3}{4}\mathbf{a}\).
Since \(Q\) is the midpoint of \(AB\), \(\vec{OQ} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\).
Therefore:
\(\vec{PQ} = \vec{OQ} - \vec{OP} = \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) - \frac{3}{4}\mathbf{a} = -\frac{1}{4}\mathbf{a} + \frac{1}{2}\mathbf{b}\).

(b)(i) Find \(\vec{OR}\):
\(\vec{OR} = \vec{OP} + \vec{PR} = \vec{OP} + 3\vec{PQ}\)
\(\vec{OR} = \frac{3}{4}\mathbf{a} + 3\left(-\frac{1}{4}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = \frac{3}{4}\mathbf{a} - \frac{3}{4}\mathbf{a} + \frac{3}{2}\mathbf{b} = \frac{3}{2}\mathbf{b}\).

(b)(ii) Since \(\vec{OR} = \frac{3}{2}\vec{OB}\), the vector \(\vec{OR}\) is a scalar multiple of \(\vec{OB}\), meaning that the point \(R\) lies on the straight line passing through \(O\) and \(B\) (or \(O, B, R\) are collinear).

(a) M1 for \(\vec{PQ} = \vec{PO} + \vec{OQ}\) or \(\vec{PA} + \vec{AQ}\) or \(\vec{OP} = \frac{3}{4}\mathbf{a}\) soi
A1 for \(-\frac{1}{4}\mathbf{a} + \frac{1}{2}\mathbf{b}\) or \(\frac{1}{2}\mathbf{b} - \frac{1}{4}\mathbf{a}\) oe in simplest form

(b)(i) B1 for \(\frac{3}{2}\mathbf{b}\) or \(1.5\mathbf{b}\)

(b)(ii) B1 for stating that \(R\) lies on the line \(OB\) / \(O, B, R\) are collinear / \(\vec{OR}\) is parallel to \(\vec{OB}\) oe

(a) Find the column vectors \(\vec{AB}\) and \(\vec{AC}\):
\(\vec{AB} = \begin{pmatrix} 5 - (-1) \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}\)
\(\vec{AC} = \begin{pmatrix} 2 - (-1) \\ 5 - 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\)

Now calculate \(\vec{AD}\):
\(\vec{AD} = 3\begin{pmatrix} 6 \\ -2 \end{pmatrix} - 2\begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 18 \\ -6 \end{pmatrix} - \begin{pmatrix} 6 \\ 4 \end{pmatrix} = \begin{pmatrix} 12 \\ -10 \end{pmatrix}\)

Find the position vector of \(D\):
\(\vec{OD} = \vec{OA} + \vec{AD} = \begin{pmatrix} -1 \\ 3 \end{pmatrix} + \begin{pmatrix} 12 \\ -10 \end{pmatrix} = \begin{pmatrix} 11 \\ -7 \end{pmatrix}\)
So the coordinates of \(D\) are \((11, -7)\).

(b) Find the vector \(\vec{CD}\):
\(\vec{CD} = \vec{OD} - \vec{OC} = \begin{pmatrix} 11 - 2 \\ -7 - 5 \end{pmatrix} = \begin{pmatrix} 9 \\ -12 \end{pmatrix}\)

Calculate the magnitude:
\(|\vec{CD}| = \sqrt{9^2 + (-12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15\).

(a) M1 for \(\vec{AB} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}\) and \(\vec{AC} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\) soi or \(\vec{AD} = \begin{pmatrix} 12 \\ -10 \end{pmatrix}\)
A1 for \((11, -7)\)

(b) M1 for \(\vec{CD} = \begin{pmatrix} 9 \\ -12 \end{pmatrix}\) soi or for \(\sqrt{(\text{their } 9)^2 + (\text{their } -12)^2}\)
A1 for 15 cao

(a) First find \(\vec{OB}\):
\(\vec{OB} = \vec{OC} + \vec{CB} = \mathbf{c} + 2\mathbf{a}\).
Since \(M\) is the midpoint of \(AB\):
\(\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB}) = \frac{1}{2}(\mathbf{a} + \mathbf{c} + 2\mathbf{a}) = \frac{1}{2}(3\mathbf{a} + \mathbf{c}) = \frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\).

(b)(i) Since \(N\) is the midpoint of \(OC\), \(\vec{ON} = \frac{1}{2}\mathbf{c}\).
\(\vec{NM} = \vec{OM} - \vec{ON} = \left(\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\right) - \frac{1}{2}\mathbf{c} = \frac{3}{2}\mathbf{a}\).

(b)(ii) \(\vec{NM} = \frac{3}{2}\vec{OA}\) shows that:
1. \(NM\) is parallel to \(OA\).
2. The length of \(NM\) is \(1.5\) times the length of \(OA\) (or \(NM : OA = 3 : 2\)).

(a) M1 for \(\vec{OB} = \mathbf{c} + 2\mathbf{a}\) soi or \(\vec{OM} = \mathbf{a} + \frac{1}{2}\vec{AB}\)
A1 for \(\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\) oe

(b)(i) B1 for \(\frac{3}{2}\mathbf{a}\) or \(1.5\mathbf{a}\)

(b)(ii) B1 for stating both: \(NM\) is parallel to \(OA\) AND \(NM\) is \(1.5\) times the length of \(OA\) (or equivalent ratio \(3:2\))

(a) \(\vec{PQ} = \vec{OQ} - \vec{OP} = \mathbf{q} - \mathbf{p}\).
Since \(PM : MQ = 2 : 3\), \(\vec{PM} = \frac{2}{5}\vec{PQ} = \frac{2}{5}(\mathbf{q} - \mathbf{p})\).
\(\vec{OM} = \vec{OP} + \vec{PM} = \mathbf{p} + \frac{2}{5}(\mathbf{q} - \mathbf{p}) = \frac{3}{5}\mathbf{p} + \frac{2}{5}\mathbf{q}\).

(b) Find vectors between pairs of the points \(K\), \(M\), and \(L\):
\(\vec{KM} = \vec{OM} - \vec{OK} = \left(\frac{3}{5}\mathbf{p} + \frac{2}{5}\mathbf{q}\right) - \frac{3}{5}\mathbf{q} = \frac{3}{5}\mathbf{p} - \frac{1}{5}\mathbf{q}\).
\(\vec{KL} = \vec{OL} - \vec{OK} = \frac{9}{5}\mathbf{p} - \frac{3}{5}\mathbf{q} = 3\left(\frac{3}{5}\mathbf{p} - \frac{1}{5}\mathbf{q}\right) = 3\vec{KM}\).
Since \(\vec{KL} = 3\vec{KM}\), \(\vec{KL}\) is parallel to \(\vec{KM}\), and because they share the common point \(K\), the points \(K\), \(M\), and \(L\) lie on a straight line.

(a) M1 for \(\vec{OM} = \mathbf{p} + \frac{2}{5}(\mathbf{q} - \mathbf{p})\) or \(\mathbf{q} - \frac{3}{5}(\mathbf{q} - \mathbf{p})\) oe
A1 for \(\frac{3}{5}\mathbf{p} + \frac{2}{5}\mathbf{q}\) or \(0.6\mathbf{p} + 0.4\mathbf{q}\)

(b) M1 for finding any valid displacement vector between the points: \(\vec{KM} = \frac{3}{5}\mathbf{p} - \frac{1}{5}\mathbf{q}\) or \(\vec{ML} = \frac{6}{5}\mathbf{p} - \frac{2}{5}\mathbf{q}\) or \(\vec{KL} = \frac{9}{5}\mathbf{p} - \frac{3}{5}\mathbf{q}\)
A1 for establishing the correct scalar relationship (e.g. \(\vec{KL} = 3\vec{KM}\) or \(\vec{ML} = 2\vec{KM}\)) AND concluding that the points lie on a straight line due to sharing a common point

(a) The modal class is the interval with the highest frequency (28):
\(45 < t \le 50\).

(b) Find the midpoints, \(x\), of each class interval:
- For \(30 < t \le 40\): midpoint \(x = 35\)
- For \(40 < t \le 45\): midpoint \(x = 42.5\)
- For \(45 < t \le 50\): midpoint \(x = 47.5\)
- For \(50 < t \le 60\): midpoint \(x = 55\)
- For \(60 < t \le 80\): midpoint \(x = 70\)

Calculate the products \(f \times x\):
- \(6 \times 35 = 210\)
- \(16 \times 42.5 = 680\)
- \(28 \times 47.5 = 1330\)
- \(22 \times 55 = 1210\)
- \(8 \times 70 = 560\)

Sum of products:
\(\sum fx = 210 + 680 + 1330 + 1210 + 560 = 3990\)

Estimate of the mean:
\(\text{Mean} = \frac{\sum fx}{\sum f} = \frac{3990}{80} = 49.875\text{ min}\) (or \(49.9\text{ min}\) correct to 3 s.f.).

(a) B1 for \(45 < t \le 50\) oe

(b) M1 for at least 4 correct midpoints seen (35, 42.5, 47.5, 55, 70)
M1 for \(\sum fx\) with their midpoints within intervals \((6 \times 35 + 16 \times 42.5 + 28 \times 47.5 + 22 \times 55 + 8 \times 70)\) soi by 3990
M1 dep on previous M1 for \(\frac{\sum fx}{80}\)
A1 for 49.875 or 49.9

(a) For the interval \(15 < x \le 25\):
- Class width = \(25 - 15 = 10\text{ cm}\).
- A class width of 10 is represented by \(2\text{ cm}\) on the drawing, so the width scale is \(\frac{2}{10} = 0.2\text{ cm}\) per unit of length.
- Frequency density (FD) = \(\frac{\text{Frequency}}{\text{Class width}} = \frac{30}{10} = 3\).
- A frequency density of 3 is represented by a height of \(6\text{ cm}\), so the vertical scale is \(\frac{6}{3} = 2\text{ cm}\) per unit of frequency density.

For the interval \(25 < x \le 40\):
- Class width = \(40 - 25 = 15\text{ cm}\).
- Width on drawing = \(15 \times 0.2 = 3\text{ cm}\).
- Frequency density = \(\frac{36}{15} = 2.4\).
- Height on drawing = \(2.4 \times 2 = 4.8\text{ cm}\).

(b) For the interval \(25 < x \le 40\):
- The interval is 15 units wide (from 25 to 40).
- The proportion with \(x > 30\) is from 30 to 40, which is \(40 - 30 = 10\) units.
- Estimated number of fish = \(\frac{10}{15} \times 36 = 24\).

For the interval \(40 < x \le 60\):
- All 35 fish have \(x > 30\).

Total estimated number of fish = \(24 + 35 = 59\).

(a) B1 for width = 3 [cm]
M1 for frequency density \(= \frac{36}{15} (= 2.4)\) or finding vertical scale factor \(2\text{ cm}\) per unit of FD soi
A1 for height = 4.8 [cm]

(b) M1 for \(\frac{10}{15} \times 36\) oe (giving 24)
A1 for 59